Ultra-QuickSort

梦里梦外; 2022-09-26 11:51 111阅读 0赞

Ultra-QuickSort

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;7000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;56247</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;20768</td> 
  </tr> 
 </tbody> 
</table>

Description

![2299_1.jpg][]In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence  
9 1 0 5 4 ,  
Ultra-QuickSort produces the output  
0 1 4 5 9 .  
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a\[i\] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
    9
    1
    0
    5
    4
    3
    1
    2
    3
    0

Sample Output

6
    0

Source

[Waterloo local 2005.02.05][]

#include<iostream>
    #include<stdio.h>
    using namespace std;
    long long coun;
    int a[1000000],temp[1000000];
    void mergearray(int a[],int first,int mid,int last)
    {
        int i=first;
        int j=mid+1;
        int m=mid;
        int n=last;
        int k=0;
        while(i<=m&&j<=n)
        {
            if(a[i]<=a[j])
            {
                temp[k++]=a[i++];
    
            }
            else
            {
                temp[k++]=a[j++];
                coun+=m-i+1;//求逆序数
            }
        }
        while(i<=m)
        {
            temp[k++]=a[i++];
        }
        while(j<=n)
        {
            temp[k++]=a[j++];
        }
        for(i=0; i<k; i++)
        {
            a[first+i]=temp[i];
        }
    }
    
    void me(int a[],int first,int last)
    {
        int mid;
        if(first<last)
        {
            mid=(first+last)/2;
            me(a,first,mid);
            me(a,mid+1,last);
            mergearray(a,first,mid,last);
    
        }
    }
    
    
    int main()
    {
    
        int n;
        int i;
        while(~scanf("%d",&n)&&n)
        {
            coun=0;
    
            for(i=0; i<n; i++)
            {
                scanf("%d",&a[i]);
            }
    
            me(a,0,n-1);
            printf("%lld\n",coun);
        }
    
    
        return 0;
    
    
    }

[2299_1.jpg]: http://poj.org/images/2299_1.jpg
[Waterloo local 2005.02.05]: http://poj.org/searchproblem?field=source&key=Waterloo+local+2005.02.05